A sphere and a cube of the same material and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The rate of heat emission by radiation is given by the Stefan-Boltzmann law: $\Delta E = e A \sigma (T^4 - T_0^4)$.Since the material,temperature,

Vedclass · Accepted Answer

$\left( \frac{\pi}{6} \right)^{1/3} : 1$

Vedclass · Answer

(C) The rate of heat emission by radiation is given by the Stefan-Boltzmann law: $\Delta E = e A \sigma (T^4 - T_0^4)$.Since the material,temperature,and surroundings are the same,the ratio of radiation emitted depends only on the surface area $A$: $\frac{(\Delta E)_{	ext{sphere}}}{(\Delta E)_{	ext{cube}}} = \frac{A_{	ext{sphere}}}{A_{	ext{cube}}} = \frac{4 \pi R^2}{6 a^2}$.Given that the volumes are equal: $\frac{4}{3} \pi R^3 = a^3$,which implies $a = R \left( \frac{4\pi}{3} ight)^{1/3}$.Substituting $a$ into the area ratio: $\frac{4 \pi R^2}{6 [R (4\pi/3)^{1/3}]^2} = \frac{4 \pi}{6 (4\pi/3)^{2/3}} = \frac{4 \pi}{6} \cdot \frac{3^{2/3}}{(4\pi)^{2/3}} = \frac{(4\pi)^{1/3}}{6 \cdot 3^{-2/3}} = \frac{(4\pi)^{1/3}}{6^{1/3} \cdot 6^{2/3} \cdot 3^{-2/3}} = \left( \frac{4\pi}{6 \cdot (4/3)^{2/3}} ight) = \left( \frac{\pi}{6} ight)^{1/3}$.Thus,the ratio is $\left( \frac{\pi}{6} ight)^{1/3} : 1$.

$A$ sphere and a cube of the same material and same volume are heated up to the same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiation emitted will be

Similar Questions

Compare the rate of loss of heat from a metal sphere at $627^{\circ} C$ with the rate of loss of heat from the same sphere at $327^{\circ} C$,if the temperature of the surrounding is $27^{\circ} C$. (nearly)

$A$ black rectangular surface of area '$A$' emits energy '$E$' per second at $27^{\circ} C$. If length and breadth are reduced to $\frac{1}{3}$ of their initial values and the temperature is raised to $327^{\circ} C$,then the energy emitted per second becomes:

Two bodies $A$ and $B$ at temperatures $T_1 \ K$ and $T_2 \ K$ respectively have the same dimensions. Their emissivities are in the ratio $16:1$. At $T_1 = x T_2$,they radiate the same amount of heat per unit area per unit time. The value of $x$ is

The surface area of a person is $1.9 \, m^2$,their body temperature is $37 \, ^oC$,and the room temperature is $22 \, ^oC$. If the temperature of the skin is $28 \, ^oC$,find the rate of heat emission. The emissivity of the skin is $0.97$.

If the sun's surface radiates heat at $6.3 \times 10^7 \ W m^{-2}$,calculate the temperature of the sun assuming it to be a black body $(\sigma = 5.7 \times 10^{-8} \ W m^{-2} K^{-4})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module